Let y=f(x) be a thrice differentiable function in (-5,5).

Question

Let $y=f(x)$ be a thrice differentiable function in $(-5,5)$. Let the tangents to the curve $y=f(x)$ at $(1, f(1))$ and $(3, f(3))$ make angles $$\pi / 6$$ and $$\pi / 4$$, respectively with positive $x$

Accepted Answer

$\begin{gathered} y=f(x)\Rightarrow \frac{dy}{dx}={f}^{'}(x) \ {\frac{dy}{dx})}_{(1,f(1))}={f}^{'}(1)=	an \frac{\pi }{6}=\frac{1}{\sqrt{3}}\Rightarrow {f}^{'}(1)=\frac{1}{\sqrt{3}} \ {\frac{dy}{dx})}_{(3,f(3))}={f}^{'}(3)=	an \frac{\pi }{4}=1\Rightarrow {f}^{'}(3)=1 \ 27\int _{1}^{3}({({f}^{'}(t))}^{2}+1){f}^{''}(t)dt=\alpha +\beta \sqrt{3} \ I=\int _{1}^{3}({({f}^{'}(t))}^{2}+1){f}^{''}(t)dt \ {f}^{'}(t)=z\Rightarrow {f}^{''}(t)dt=dz \ z={f}^{'}(3)=1 \ z={f}^{'}(1)=\frac{1}{\sqrt{3}} \end{gathered}$

$\begin{gathered} I=\int _{1/\sqrt{3}}^{1}({z}^{2}+1)dz={(\frac{{z}^{3}}{3}+z)}_{1}^{1/\sqrt{3}} \ =(\frac{1}{3}+1)-(\frac{1}{3}\cdot \frac{1}{3\sqrt{3}}+\frac{1}{\sqrt{3}}) \ =\frac{4}{3}-\frac{10}{9\sqrt{3}}=\frac{4}{3}-\frac{10}{27}\sqrt{3} \ \alpha +\beta \sqrt{3}=27(\frac{4}{3}-\frac{10}{27}\sqrt{3})=36-10\sqrt{3} \ \alpha =36,\beta =-10 \ \alpha +\beta =36-10=26 \end{gathered}$

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